91Ó°ÊÓ

A triangle is given.

A triangle congruent to $\mathrm{Δ}OBT$ using $ASA$ postulate,

Consider the following ways-

Draw a line $L$.

Mark a point $F$on the line $L$.

Taking $F$as center draw an arc on the line $L$with radius of $BT$.

With $B$as center draw an arc which touches the sides $BO$ and $BT$.

With same radius draw an arc with $F$as center.

Then mark the distance between the points $P$ and $Q$.

Draw an arc of same measure with $S$ as center.

Draw a ray at the point of intersection, marked it as $J$.

With $T$as center, draw an arc which touches the sides $TB$ and $TO$.

With same radius draw an arc with $G$as center.

Then mark the distance between the points $U$ and $V$.

Draw an arc of same measure with $X$as center.

Draw a ray at the point of intersection, marked it as $K$.

Then mark the point of intersection as $E$.

Angle-Side-Angle theorem states that,

if two angles and any side of one triangle are congruent to two angles and any side of another triangle,

then the triangles are congruent.

Thus, $\mathrm{Δ}OBT$ is congruent to $\mathrm{Δ}EFG$.

Since, $∠Bâ‰\dots ∠F,∠Tâ‰\dots ∠G$ and $BTâ‰\dots FG$

then $\mathrm{Δ}OBTâ‰\dots \mathrm{Δ}EFG$.

Therefore, by $ASA$ method $\mathrm{Δ}OBTâ‰\dots \mathrm{Δ}EFG$.